$f\colon \Bbb R^3 \to \Bbb R^3 $ be defined by $f(x_1,x_2,x_3)=....$ I am stuck on the following problem:  

Let $f\colon \Bbb R^3 \to \Bbb R^3 $ be defined by $f(x_1,x_2,x_3)=(x_2+x_3,x_3+x_1,x_1+x_2).$ Then the first derivative of $f$ is :
  1.not invertible anywhere
  2.invertible only at the origin
  3.invertible everywhere except at the origin
  4.invertible everywhere.   

My problem is I do not know how to calculate the derivative of $f$. Can someone point me in the right direction?
 A: HINT: the derivative is a linear transformation $f':\mathbb{R^3}\rightarrow\mathbb{R}^3$ such that its matrix is
$$[f']=\left( \begin{array}{ccc}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} \\
\frac{\partial f_3}{\partial x_1} & \frac{\partial f_3}{\partial x_2} & \frac{\partial f_3}{\partial x_3}
\end{array} \right)$$
A: The question is most likely looking for the Jacobian:
$$J=f'(x_1,x_2,x_3)=\begin{bmatrix}\frac{\partial f_1}{\partial{x_1}} & \frac{\partial f_1}{\partial{x_2}} & \frac{\partial f_1}{\partial{x_3}}\\\frac{\partial f_2}{\partial{x_1}} & \frac{\partial f_2}{\partial{x_2}} &\frac{\partial f_2}{\partial{x_1}}\\\frac{\partial f_3}{\partial{x_1}} & \frac{\partial f_3}{\partial{x_2}} &\frac{\partial f_3}{\partial{x_3}}\end{bmatrix}$$
This seems like a lot of work, but since this is a linear transformation, the Jacobian will be identical to the transformation matrix itself.
A: The derivative of $f$ will be a three-by-three matrix, where the mnth entry is $\frac{\partial f_m}{\partial x_n}$. 
This means row 1 column 1 entry will be 0, row 1 column 2 will be 1. row 1 column 3 entry will be 1 etc.
